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Annette [7]
2 years ago
9

An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

s at the 16 inch mark, crawls to the 29-inch mark, then moves to the 14-inch mark. What is the total distance the ant traveled? What was the total displacement of the ant?
Physics
1 answer:
FrozenT [24]2 years ago
6 0

Answer:

  • The total distance traveled is 28 inches.
  • The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector \hat{i} pointing in the west direction, the ant start at position

\vec{r}_0 = 16 \ inch \ \hat{i}

Then, moves to

\vec{r}_1 = 29 \ inch \ \hat{i}

so, the distance traveled here is

d_1 = |\vec{r}_1 - \vec{r}_0  | = | 29 \ inch   \ \hat{i} - 16 \ inch   \ \hat{i}  |

d_1 =  | 13 \ inch   \ \hat{i}  |

d_1 =  13 \ inch

after this, the ant travels to

\vec{r}_2 = 14 \ inch \ \hat{i}

so, the distance traveled here is

d_2 = |\vec{r}_2 - \vec{r}_1  | = | 14 \ inch   \ \hat{i} - 29 \ inch   \ \hat{i}  |

d_2 =  | - 15 \ inch   \ \hat{i}  |

d_2 =  15 \ inch

The total distance traveled will be:

d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch

The displacement is the final position vector minus the initial position vector:

\vec{D}=\vec{r}_2 - \vec{r}_1

\vec{D}= 14 \ inch   \ \hat{i} - 16 \ inch \ \hat{i}

\vec{D}= - 2 \ inch \ \hat{i}

This is 2 inches to the east.

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ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
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By definition, the moment of inertia in positions 1 and 2 are respectively
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Because momentum is conserved,
I₁ω₁ = I₂ω₂
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ω₂ = (I₁/I₂)ω₁
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The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
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F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
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Answer:  44.25 ft-lb


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Explanation:

<em><u>The knowable variables are </u></em>

d_{t}=25m

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t_{3}=5.5 s

Since the three traffic signs are <u>equally spaced</u>, the <u>distance between each sign is \frac{25}{3} m</u>

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